# Prove that:

Question:

Prove that:

$\left|\begin{array}{lll}1 & a & b c \\ 1 & b & c a \\ 1 & c & a b\end{array}\right|=\left|\begin{array}{lll}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{array}\right|$

Solution:

Let $\quad$ LHS $=\Delta=\mid \begin{array}{lll}1 & a & b c\end{array}$

$\begin{array}{lll}1 & b & c a \\ 1 & c & a b\end{array}$

$=\frac{1}{\mathrm{abc}} \mid \begin{array}{lll}a & a^{2} & a b c\end{array}$

$b \quad b^{2} \quad b c a$

$c \quad c^{2} \quad a b c \mid \quad$ [Applying $\mathrm{R}_{1} \rightarrow \mathrm{a} \mathrm{R}_{1}, \mathrm{R}_{2} \rightarrow \mathrm{b} \mathrm{R}_{2}$ and $\mathrm{R}_{3} \rightarrow \mathrm{c} \mathrm{R}_{3}$ and then dividing it by $a b c$ ]

$=\frac{a b c}{a b c} \mid a \quad a^{2} \quad 1$

$\begin{array}{llll}b & b^{2} & 1 & \\ c & c^{2} & 1 \mid & \text { [Taking out } a b c \text { common from } \mathrm{C}_{3} \text { ] }\end{array}$

$=(-1) \mid 1 \quad a^{2} \quad a$

$\begin{array}{ccc}1 & b^{2} & b \\ 1 & c^{2} & c\end{array}$

[Interchanging $\mathrm{C}_{3}$ and $\mathrm{C}_{1}$ to get $-$ ve value of original determinant]

$=(-1)(-1) \mid 1 \quad a \quad a^{2}$

$\begin{array}{cccc}1 & b & b^{2} & \\ 1 & c & c^{2} & {\left[\text { Applying } \mathrm{C}_{2} \leftrightarrow \mathrm{C}_{3}\right]}\end{array}$

$\begin{array}{cccc}= & \mid 1 & a & a^{2} \\ 1 & b & b^{2} & \\ 1 & c & c \mid & \end{array}$

$=\mathrm{RHS}$